Study of different flows over typical bodies by Fluent

Study of different flows over typical bodies by Fluent

Under the supervision ofDr. M.K. Laha

RAJIBUL ALAM14AE60R03

DEPARTMENT OF AEROSPACE ENGINEERINGIIT KHARAGPUR

Inviscid flow over a wedge

Inviscid flow over a wedge is governed by continuity, momentum and energy equations which are given as follows.

Continuity equation:

Momentum equation:

Energy equation:

0.dSVdvt

0. Vt

fdvpdSVdSVVdvt

).(

zfzp

DtDw

yfyp

DtDv

xfxp

DtDu

dvVfdSpVdvqdSVVedvVet

).(..

.)22

()2

2(

).(..

)2

2( VfpVqVe

DtD

Assumptions:

Flow is steady Flow is adiabatic There are no viscous effects on the sides of the control volume which includes the

shock wave. There is no body force.

With above assumptions, the normal shock equations become as follows:

Continuity:

Momentum:

Energy:

Thus there are four unknowns. With addition of following thermodynamic relations, system becomes of 5 equations and 5 unknowns.

,

2211 uu

upup 222111

2

22

22

21

1V

hV

h

22 Tpch RTp 22

On solving the above system, we have following relations:

2)2cos(21

12sin21cot2tan

M

M

2/)1(2sin21

2sin21]2/)1[(1

)(2sin22

M

MM

)12sin21(

121

1

2

Mp

p

2sin2

1)1(2

2sin21)1(

1

2M

M

2

1

1

2

1

2

p

p

T

T

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Now a the given problem,

With the previous relations, we can determine,

Numerical Solution for the Given Problem:

Fluent is used for a numerical solution.

Flow field dimensions are shown below.

o15KTatmpM 3001,11,31

2968.22 M atmp 78.22 477.4162 T

1.259m

1.5m

0.991m

0.5m

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Meshing:

Faced mapping is used which create quadrilateral meshing elements. The size of each element is taken as 0.05m. Also density based approach is used which is suitable for supersonic flow.

Material:

Fluid is assumed as ideal gas with specific heat constant 1000.43 J/Kg.K and molecular weight 28.966 Kg/mol. The wedge material is assumed as Aluminum.

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Results:

Mach variation:

3,15 M

Numerical vs. Analytical :

4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00 1.40E+00 1.60E+000.00E+00

5.00E-01

1.00E+00

1.50E+00

2.00E+00

2.50E+00

3.00E+00

3.50E+00

Mach Number Comparision for M=3

Numerical Analytical

3,15 M

Pressure variation:

3,15 M


: Numerical

: Analytical

4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00 1.40E+00 1.60E+000.00E+00

5.00E+04

1.00E+05

1.50E+05

2.00E+05

2.50E+05

3.00E+05

3.50E+05

3,15 M

Pressure coefficient:

3,15 M


4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00 1.40E+00 1.60E+000.00E+00

5.00E-02

1.00E-01

1.50E-01

2.00E-01

2.50E-01

3.00E-01

3.50E-01

3,15 M

Temperature variation:

3,15 M

4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00 1.40E+00 1.60E+000.00E+00

5.00E+01

1.00E+02

1.50E+02

2.00E+02

2.50E+02

3.00E+02

3.50E+02

4.00E+02

4.50E+02


MESH REFINEMENT:

It has been seen that changes across the shock wave is gradual whether in real cases, such changes are instantaneous. For a better picture of the situation, mesh refinement is done.

In the earlier case, simulation was done with 713 meshing elements. For mesh refinement purpose, number of elements is increased to 90,000. Following are the results obtained.

Mach variation:

3,15 M


3,15 M

Pressure variation:

3,15 M

3,15 M

Numerical vs. Analytical:


3,15 M

3,15 M

Numerical vs. Analytical

Temperature:

3,15 M


3,15 M

Solution with increased Mach number: Now the free stream Mach number is increased to 5 keeping the geometry same and numerical solution is obtained with same type of meshing and size as that of for M=3.

• Analytical solution:

• M=3.5,

Numerical solution:

Mach Number:

atmp 78.42 KT 8.5202

5,15 M


5,15 M

Pressure:

5,15 M


5,15 M


5,15 M


5,15 M

Temperature

5,15 M


5,15 M

Viscous Flow over a Flat Plate

Boundary Layer Equations:

With boundary layer approximation, we have,

X-momentum equation as,

Y-momentum equation:

xp

yu

yuv

xuu

2

2

)(

yp

xv

yvv

xvu

2

2

)(

Assumptions:

v<<u, yu

xu

yv

xv

,0,0

yT

xT

Applying these assumptions, we have,

0yp

Now for a flat plate, since u= =CONSTANT and v=0 outside the boundary layer, hence X-momentum equation gives, 0

xp

Thus boundary layer equations become,

0

yv

xu

2

2

yu

yuv

xuu

2

2

yT

yT

xTu

Continuity

Momentum

Energy

Analytical Solution Procedure:

• Blasius solved continuity and momentum equations by transforming the equations into a single ordinary differential equation.

• He did it by introducing a new independent variable called similarity variable.

• Blasius reasoned that when non dimensional velocity is plotted against non dimensional distance, it gives same variation at any point along the plate.

• Now introducing,

the two differential equations can be transformed into a single ordinary differential equation as,

y

)(

gUu

dgf )(

02 2

2

3

3

dfdf

dfd

It can be shown that u= and ddfU )(

21 f

ddf

xU

v

Now above ODE is solved by Runge Kutta method with known boundary equations.

On solving, following values are obtained.

f

0 0 0 0.332

0.5 0.042 0.166 0.331

1 0.166 0.33 0.323

1.5 0.37 0.487 0.303

2 0.65 0.63 0.267

2.5 0.996 0.751 0.217

3 1.397 0.846 0.161

3.5 1.838 0.913 0.108

4 2.306 0.956 0.064

4.5 2.79 0.980 0.034

5 3.283 0.992 0.016

5.5 3.781 0.997 0.007

6 4.28 0.999 0.002

1 0

Uu

ddf

2

2

dfd

Thus when

Ux

Uxy

Uu //0.5,992.0

xU

/5

xeyw R

Udfd

xUU

yu 2

02

2

0332.0

xexf RC /664.0,

This gives,

Similarly,

Local skin friction coefficient becomes,

Numerical solution:

Geometry and flow field:

Flat plate is considered as two dimensional one with a length of 1m. The flow field and boundaries are shown in the following figure.

Meshing:

Mapped face meshing is applied.

The upper and lower edges are given 500 divisions with no biasing.

On the other hand, the left and right side edges are given 500 numbers of divisions and biasing with biasing factor of 200.

There are 250000 elements with 250001 nodes.

Set up:

Since the flow is incompressible one, hence pressure based solver is used.

Energy solution is kept off and viscous laminar type flow is chosen.

Boundary conditions:

The far field edge is given as symmetry boundary.

Inlet is given as velocity inlet boundary condition with inlet velocity 1m/s .

Plate is given as wall boundary condition while outlet is given as pressure outlet with gauge pressure=0.

Results:

Normalized velocity vs. normalized distance at quarter chord point

Normalized velocity vs. normalized distance at half chord point

Normalized velocity vs. normalized distance at three quarter chord point

Normalized velocity vs. normalized distance at the edge of the plate

Now all the four plots are taken on the same plane to prove Blasius assumption.

Wall shear stress distribution:

Numerical vs. Analytical:

SUPERSONIC FLOW OVER A CONE

First we will consider a general cone with z-axis as the axis of symmetry and aligned with the axis of the cone. Also free stream velocity is this direction i.e. along the axis of the cone.

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Formulation:

Now since body is axisymmetric, hence

Since properties are constant along any ray, hence

0

0r

Now continuity equation is,0)(

sin1)sin(

sin1)(1 2

2

Vr

Vr

Vrrr r

On simplification, 0cot2

VVVVr


When Bernoulli’s equation is associated with the above equation, it gives Taylor-McColl equation given by,

0))(()2)()((2

)1(2

2

2

2222

max

dVd

ddV

ddVV

ddV

dVd

ddVV

ddVVV rrr

rrrr

rr

r

Numerical solution:

A numerical solution of a cone of base radius 0.1 m is obtained in fluent.

Free stream Mach number is 2 and pressure is 101325 Pa.

Cone geometry is created through primitive geometry.

Flow field is chosen as rectangular parallelepiped of sides 10 m symmetrical about the coordinate axes.

Meshing:

All the surfaces of the parallelepiped are named as far field boundary and cone surface is named as cone surface.

Default meshing is generated with 131567 elements.

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Solution:

In the CFD post, pressure, temperature and density are plotted along a ray with coordinates (0, 0, 0) and (1, 1, 1) and following results are obtained.


Pressure variation along the specified ray:

Density along the given specified ray:

Temperature distribution along a given ray:

Flow past a sphere at low Reynold’s number (Creeping flow)

Creeping flow is the one that flows past a small sphere at very low Reynold’s number. Stoke gave solution for such a flow.

Relevant Equations:

Consider a flow past a sphere as shown in figure. Let the flow is in Z-direction. Due to axial symmetry,

0

0 V

0sin1)sin(1)(1 2

2

V

rV

rVr

rr r

0)sin(1)(1 22

VrVr

rr r

Continuity:

i.e.

(1)

Momentum equation:

Neglecting inertia, 02

Vp

cosVVr

sinVV

Now, at infinity, ,

These give, rrV ,sin2

22

Now associating this with continuity, momentum gives,

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Drag:

Radial strain is given by,

On the sphere surface, r=a. Hence,

Hence,

Tangential strain is given by,

At r=a,

Thus, total stress in Z direction= =

)23

23(cos

21

4

3

2 ra

raV

rVe r

rr

0rre

prr

sin

23)1)( 4

3

rVaV

rrV

rre r

r

sin23aVer

sincos rrr aVp 2

3cos

The constant part gives a net drag in Z direction which can be determined as,

D= Va6

deD RC 24

Numerical solution for an inviscid flow past a small sphere at low Reynold’s number:

Geometry:A sphere of 0.1m radius is considered in a free stream of velocity 1 m/s. A spherical geometry is created through primitive geometry.

Meshing:

Default meshing is created with 65502 elements. Inlet and outlet surfaces are named as inlet and outlet while other surfaces are named as boundary wall. Sphere wall is named as sphere wall.

Set up:

Inlet is given as velocity inlet boundary condition, outlet is given as pressure outlet boundary condition while all other surfaces are given as symmetry boundary condition.

Result:

Plots from CFD post for drag and lift coefficients are obtained.

Iteration Continuity x-velocity y-velocity z-velocity Energy265 8.9615e-

081.0145e-06

4.7521e-07

9.6569e-07

1.3218e-16

9.0248e-07

5.2368e-06

266 8.8382e-08

1.0089e-06

4.6785e-07

9.6178e-07

1.3117e-16

9.1195e-7

5.2367e-6

267 8.8382e-08

1.0046e-06

4.6123e-07

9.5837e-07

1.3274e-16

9.1759e-07

5.2384e-06

LC DC

Remarks: From Stoke’s theorem as well from symmetry, it is evident that drag as well lift are zero and hence their coefficients. This is obtained from the numerical solution too.

Convergent results table:

References: Modern Compressible Flow- John D Anderson

Viscous Fluid Flow- Frank M White

Notes On Advanced Environment Fluid Mechanics- C.C. Mei

Introduction To Aerodynamics- John D Anderson

Engineering

Study of different flows over typical bodies by Fluent